Let
\[f(x) = \left\{
\begin{array}{cl}
x + 3 & \text{if $x < 20$}, \\
2x - 2 & \text{if $x \ge 20$}.
\end{array}
\right.\]Find $f^{-1}(7) + f^{-1}(46).$
To find $f^{-1}(7),$ we try solving $f(x) = 7$ on each piece.

If $x + 3 = 7$, then $x = 4,$ which satisfies $x < 20.$  If $2x - 2 = 7,$ then $x = \frac{9}{2},$ which does not satisfy $x \ge 20,$ so $f^{-1}(7) = 4.$

Similarly, $x + 3 = 46,$ then $x = 43,$ which does not satisfy $x < 20.$  If $2x - 2= 46,$ then $x = 24,$ which satisfies $x \ge 20,$ so $f^{-1}(46) = 24.$

Hence, $f^{-1}(7) + f^{-1}(46) = 4 + 24 = \boxed{28}.$